1. Key Concepts and Formulas

This problem involves two fundamental concepts from 3D Geometry:

• Centroid of a Triangle: For a triangle with vertices $A(x_1, y_1, z_1)$, $B(x_2, y_2, z_2)$, and $C(x_3, y_3, z_3)$, its centroid $G$ is given by the average of the coordinates: $G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$
• Image of a Point in a Plane: The image $P'(\alpha, \beta, \gamma)$ of a point $P(x_1, y_1, z_1)$ in a plane $ax+by+cz+d=0$ is found using the formula: $\frac{\alpha - x_1}{a} = \frac{\beta - y_1}{b} = \frac{\gamma - z_1}{c} = -2 \frac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}$ This formula efficiently combines the conditions that the line segment $PP'$ is perpendicular to the plane and its midpoint lies on the plane.

2. Step-by-Step Solution

Step 1: Finding the Coordinates of the Third Vertex C

• What we are doing: We are using the centroid formula to find the coordinates of the unknown third vertex C.
• Why we are doing this: The problem asks for the image of the third vertex. Since only two vertices are given, we must first determine the coordinates of this third vertex.

We are given:

• Vertex $A = (2, 4, 6)$
• Vertex $B = (0, -2, -5)$
• Centroid $G = (2, 1, -1)$ Let the third vertex be $C = (x_C, y_C, z_C)$.

Using the centroid formula: $G = \left(\frac{x_A+x_B+x_C}{3}, \frac{y_A+y_B+y_C}{3}, \frac{z_A+z_B+z_C}{3}\right)$ Substitute the given coordinates: $(2, 1, -1) = \left(\frac{2+0+x_C}{3}, \frac{4+(-2)+y_C}{3}, \frac{6+(-5)+z_C}{3}\right)$ $(2, 1, -1) = \left(\frac{2+x_C}{3}, \frac{2+y_C}{3}, \frac{1+z_C}{3}\right)$

Now, equate the corresponding coordinates to solve for $x_C, y_C, z_C$:

• For the x-coordinate: $\frac{2+x_C}{3} = 2 \implies 2+x_C = 6 \implies x_C = 4$
• For the y-coordinate: $\frac{2+y_C}{3} = 1 \implies 2+y_C = 3 \implies y_C = 1$
• For the z-coordinate: $\frac{1+z_C}{3} = -1 \implies 1+z_C = -3 \implies z_C = -4$ Thus, the coordinates of the third vertex are $C(4, 1, -4)$.

Step 2: Determining the Image of Vertex C in the Given Plane

• What we are doing: We are applying the formula for the image of a point in a plane to find the coordinates of the reflection of vertex C.
• Why we are doing this: The problem specifically asks for an expression involving the coordinates of the image of the third vertex. This step directly calculates those coordinates.

We need to find the image of $C(4, 1, -4)$ in the plane $x+2y+4z=11$. Let the image be $D(\alpha, \beta, \gamma)$. First, write the plane equation in the standard form $ax+by+cz+d=0$: $x+2y+4z-11=0$ From this, we identify the coefficients: $a=1, b=2, c=4, d=-11$. The point is $C(x_1, y_1, z_1) = (4, 1, -4)$.

Now, apply the formula for the image of a point in a plane: $\frac{\alpha - x_1}{a} = \frac{\beta - y_1}{b} = \frac{\gamma - z_1}{c} = -2 \frac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}$ Substitute the values: $\frac{\alpha - 4}{1} = \frac{\beta - 1}{2} = \frac{\gamma - (-4)}{4} = -2 \frac{1(4)+2(1)+4(-4)-11}{1^2+2^2+4^2}$

Calculate the numerator and denominator of the fraction on the right side:

• Numerator: $1(4)+2(1)+4(-4)-11 = 4+2-16-11 = 6-27 = -21$
• Denominator: $1^2+2^2+4^2 = 1+4+16 = 21$

Substitute these values back into the formula: $\frac{\alpha - 4}{1} = \frac{\beta - 1}{2} = \frac{\gamma + 4}{4} = -2 \frac{-21}{21}$ $\frac{\alpha - 4}{1} = \frac{\beta - 1}{2} = \frac{\gamma + 4}{4} = -2(-1)$ $\frac{\alpha - 4}{1} = \frac{\beta - 1}{2} = \frac{\gamma + 4}{4} = 2$

Now, equate each part to 2 to find $\alpha, \beta, \gamma$:

• For $\alpha$: $\frac{\alpha - 4}{1} = 2 \implies \alpha - 4 = 2 \implies \alpha = 6$
• For $\beta$: $\frac{\beta - 1}{2} = 2 \implies \beta - 1 = 4 \implies \beta = 5$
• For $\gamma$: $\frac{\gamma + 4}{4} = 2 \implies \gamma + 4 = 8 \implies \gamma = 4$ So, the coordinates of the image of the third vertex are $D(\alpha, \beta, \gamma) = (6, 5, 4)$.

Step 3: Calculating the Required Expression

• What we are doing: We are substituting the derived coordinates of the image into the expression requested by the problem.
• Why we are doing this: This is the final calculation required to answer the question.

We need to calculate $\alpha\beta+\beta\gamma+\gamma\alpha$. Using the values we found: $\alpha=6, \beta=5, \gamma=4$.

Substitute these values into the expression: $\alpha\beta+\beta\gamma+\gamma\alpha = (6)(5) + (5)(4) + (4)(6)$ $= 30 + 20 + 24$ $= 74$

3. Common Mistakes & Tips

• Centroid Calculation: Be careful with signs, especially when subtracting negative coordinates. A common mistake is to forget that $x_A+x_B+x_C = 3G_x$.
• Plane Equation Form: Always rewrite the plane equation into the standard form $ax+by+cz+d=0$ to correctly identify the value of $d$ (including its sign).
• Image Formula Sign: Remember the factor of $-2$ in the image formula. Using $-1$ would give the foot of the perpendicular from the point to the plane.
• Arithmetic Precision: Double-check all arithmetic, particularly when dealing with fractions or multiple negative signs. These problems often have integer answers, so non-integer intermediate results for $\alpha, \beta, \gamma$ might indicate an error.

4. Summary

We systematically solved this problem by first determining the coordinates of the third vertex using the centroid formula. Then, we applied the standard formula for the image of a point in a plane to find the coordinates of its reflection. Finally, we calculated the required algebraic expression using these coordinates. The resulting value for $\alpha\beta+\beta\gamma+\gamma\alpha$ is 74.

5. Final Answer

The final answer is $\boxed{74}$, which corresponds to option (B).